Question

Decomposing vectors For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). $$\mathbf{u}=\langle-1,2,3\rangle, \mathbf{v}=\langle 2,1,1\rangle$$

Short answer

Question: Express the vector \(\mathbf{u} = \langle -1, 2, 3 \rangle\) as a sum of two vectors \(\mathbf{p}\) and \(\mathbf{n}\) such that \(\mathbf{p}\) is parallel to \(\mathbf{v} = \langle 2, 1, 1 \rangle\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). Answer: \(\mathbf{u} = \mathbf{p} + \mathbf{n} = \langle 1, 0.5, 0.5 \rangle + \langle -2, 1.5, 2.5 \rangle\).

Step-by-step solution

Find the component of \(\mathbf{u}\) parallel to \(\mathbf{v}\)

To find the component of \(\mathbf{u}\) that is parallel to \(\mathbf{v}\), first check the dot product of both vectors and then multiply the resulting scalar by the unit vector of \(\mathbf{v}\). So, $$\mathbf{p} = \frac{\mathbf{u} \cdot \mathbf{v}}{ ||\mathbf{v}||^2} \times \mathbf{v}$$ First, we calculate the dot product of \(\mathbf{u}\) and \(\mathbf{v}\): $$\mathbf{u} \cdot \mathbf{v} = (-1)(2) + (2)(1) + (3)(1) = -2 + 2 + 3 = 3$$ Now, we find the square of the magnitude of \(\mathbf{v}\): $$||\mathbf{v}||^2 = 2^2 + 1^2 + 1^2 = 6$$ Finally, we calculate the parallel component \(\mathbf{p}\): $$\mathbf{p} = \frac{3}{6} \times \langle 2, 1, 1 \rangle = \frac{1}{2} \times \langle 2, 1, 1 \rangle = \langle 1, 0.5, 0.5 \rangle$$

Find the component of \(\mathbf{u}\) orthogonal to \(\mathbf{v}\)

To find the component of \(\mathbf{u}\) that is orthogonal to \(\mathbf{v}\), simply subtract the parallel component \(\mathbf{p}\) from \(\mathbf{u}\): $$\mathbf{n} = \mathbf{u} - \mathbf{p}$$ $$\mathbf{n} = \langle -1, 2, 3 \rangle - \langle 1, 0.5, 0.5 \rangle = \langle -2, 1.5, 2.5 \rangle$$

Express \(\mathbf{u}\) as the sum of \(\mathbf{p}\) and \(\mathbf{n}\)

Now that we have found both \(\mathbf{p}\) and \(\mathbf{n}\), we can express \(\mathbf{u}\) as the sum of these two vectors: $$\mathbf{u} = \mathbf{p} + \mathbf{n} = \langle 1, 0.5, 0.5 \rangle + \langle -2, 1.5, 2.5 \rangle = \langle -1, 2, 3 \rangle$$

Key concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It allows us to measure how much two vectors share in common direction. Let's consider two vectors, \( \mathbf{a} \) and \( \mathbf{b} \). The dot product is computed as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \) for 3-dimensional vectors
• The result is a scalar, not a vector

In the original problem, the dot product of vectors \( \mathbf{u} = \langle -1, 2, 3 \rangle \) and \( \mathbf{v} = \langle 2, 1, 1 \rangle \) is calculated as \( -2 + 2 + 3 = 3 \). This scalar result is crucial in vector decomposition.

Why is it crucial? The dot product helps us find a vector’s projection in the direction of another vector. It is a means to extract the part of one vector that runs along the line of the other vector. If the dot product is zero, this indicates that the vectors are perpendicular.

Parallel Vectors

When dealing with vectors, being parallel means one vector is a scalar multiple of the other. This concept is useful in many mathematical applications including physics and engineering. In the context of our original exercise, the component of vector \( \mathbf{u} \) that is parallel to vector \( \mathbf{v} \) is obtained by the projection formula:

• \( \mathbf{p} = \frac{\mathbf{u} \cdot \mathbf{v}}{ ||\mathbf{v}||^2} \times \mathbf{v} \)
• \( ||\mathbf{v}||^2 \) is the square of the magnitude of vector \( \mathbf{v} \)

In simpler terms, you first scale the vector \( \mathbf{v} \) to become a unit vector, then scale it again by the scalar dot product value.

This makes sure \( \mathbf{p} \) is both parallel to \( \mathbf{v} \) and captures exactly how much of \( \mathbf{u} \) aligns with \( \mathbf{v} \). For our problem, the parallel component \( \mathbf{p} \) was \( \langle 1, 0.5, 0.5 \rangle \), achieving both properties: aligning directionally with \( \mathbf{v} \) and sharing the appropriate magnitude.

Orthogonal Vectors

Orthogonal vectors are vectors that are at a right angle (90 degrees) to each other. Whenever the dot product of two vectors is zero, these vectors are orthogonal. In our vector decomposition, discovering a component of \( \mathbf{u} \) orthogonal to \( \mathbf{v} \) means finding any part of \( \mathbf{u} \) that diverges completely from \( \mathbf{v} \).

To find the orthogonal vector \( \mathbf{n} \), we subtract the parallel component \( \mathbf{p} \) from \( \mathbf{u} \). This yields

• \( \mathbf{n} = \mathbf{u} - \mathbf{p} \)
• In our example: \( \mathbf{n} = \langle -2, 1.5, 2.5 \rangle \)

No part of the orthogonal component \( \mathbf{n} \) points in the direction of \( \mathbf{v} \). This scenario highlights their zero-dot product relationship.

The orthogonal component is significant because it completely captures the aspect of one vector that is perpendicular to the other, thus giving full insight into how vectors can be analyzed and understood in relation to each other in space.